Example Questions

Example Question #1 : Heat Transfer And Thermal Equilibrium

of energy is applied to an aluminum rod of unknown mass. Its temperature goes from to . What is the mass of the rod?

Possible Answers:

There is not enough information to determine the rod's mass

Correct answer:

Explanation:

The relevant equation for this problem is called the specific heat capacity equation:

In this equation, is the total energy in Joules, is the mass in grams, is the specific heat of the substance in Joules over grams times Coulombs , and is the change in temperature in Kelvins or degrees Celsius; which one you use doesn't matter because it's the change you need.

To find the mass of the rod when given the energy, specific heat, and change in temperature, we can rearrange the equation to this:

Example Question #1 : Heat Transfer And Thermal Equilibrium

Which of the following best defines the zeroth law of thermodynamics in variable form?

Possible Answers:

Correct answer:

Explanation:

The zeroth law of thermodynamics states that if an object X is in thermal equilibrium with object Y, and object Y also in thermal equilibrium with object Z, then object X must also be in thermal equilibrium with object Z.

Example Question #3 : Heat Transfer And Thermal Equilibrium

You have of water in a container above a burner. If of energy is put into the container (assume all of it goes into the water), and the specific heat of water is , how much did the water's temperature rise?

Possible Answers:

There's not enough information to determine the temperature change

Correct answer:

Explanation:

The equation for temperature change given applied heat is

.

is the amount of heat energy, is the mass, is the specific heat, and is the change in temperature. In this problem, we're given the energy, the mass, and the specific heat, so we need to solve the equation for .

Rearranging to find the change in temperature:

Now, we can plug in our numbers.

Therefore, the water changed by about degrees celsius given that amount of energy applied.

Example Question #5 : Heat Transfer And Thermal Equilibrium

Example Question #6 : Heat Transfer And Thermal Equilibrium

The heat capacity for any given material is defined as the amount of heat that must be added to that material in order to increase its temperature by . However, the heat capacity for any given material can be measured under different conditions, such as constant pressure or constant volume.

How would the heat capacity for a material at constant volume differ from the heat capacity for that same material at constant pressure?

Possible Answers:

The heat capacity at constant volume is greater than the heat capacity at constant pressure

There is no way to tell; the relative values of the constant volume and constant pressure heat capacities depends on the material under consideration

The heat capacity at constant pressure is greater than the heat capacity at constant volume

The heat capacity at constant volume is equal to the heat capacity at constant pressure

Correct answer:

The heat capacity at constant pressure is greater than the heat capacity at constant volume

Explanation:

In this question, we're presented with the definition of heat capacity. We're also told that the heat capacity for any given material can be measured at 1) constant volume, or 2) constant pressure. We're being asked to determine whether the constant volume heat capacity is greater than, less than, or equal to the constant pressure heat capacity for the same material.

First, let's revisit the definition of heat capacity. It is the amount of heat that must be added to a material to raise the temperature of that material by .

Next, let's consider what is happening as we add heat to a material at constant volume. When the volume of a material is held constant, that means that it is prevented from expanding as heat is added. Thus, all of the heat being added is resulting directly in an increase in the average kinetic energy of the particles that make up that material. In other words, all of the heat being added results in a temperature increase.

Let's now consider what happens when heat is added to a material at constant pressure. In this situation, the material is allowed to expand as heat is added to it (because volume is not being held constant). Therefore, when heat is added to the material, some of the heat will result in an increased temperature, but some of the heat will also result in an expansion of the material. The consequence of this is that in order to increase the temperature of the material by a given amount at constant pressure, it will require more heat energy than at constant volume.

Again, the main point of this is as follows:

Constant volume - all of the added heat results in increased temperature

Constant pressure - some of the added heat results in increased temperature and some of the added heat results in expansion

Since only a fraction of the heat being added at constant pressure will actually result in a temperature increase, we'll have to add more heat in order to increase its temperature by a given amount, when compared to constant volume conditions.

Example Question #7 : Heat Transfer And Thermal Equilibrium

of ethanol is at a temperature of and of methanol is at a temperature of . If the two gases are in an isolated system together and allowed to transfer heat between each other, at which temperature will they reach thermal equilibrium?

Possible Answers:

Correct answer:

Explanation:

Since the two gases are in an isolated system, we know that the heat lost by one is equal to the heat gained by the other. Furthermore, we know that the two gases will be in thermal equilibrium when they have the same temperature, which will be some value in between their two initial values. Since methanol is at a higher temperature, we can say that it will loose heat and ethanol will gain that same amount of heat:

Example Question #10 : Heat Transfer And Thermal Equilibrium

An ideal heat engine uses the atmosphere as a low temperature reservoir and boiling water as a hot temperature reservoir. Water boils at . How much more efficient is the heat engine on a cold day () than when it is hot outside ()?

Possible Answers:

times more efficient

Not enough information.

times more efficient

Equally efficient

times more efficient

Correct answer:

times more efficient

Explanation:

The efficiency of an ideal heat engine is given by:

, where and stand for hot and cold respectively. To use this equation our temperatures must be in Kelvin. Here is the equation to convert from Celsius to Kelvin.

.

The denominator does not change if the temperature is hot or cold, so we need only consider the numerator. Let's compare both calculations:

Hot Day:

Cold Day:

Comparing these two efficiencies, we see that the cold day is roughly times as efficient.

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